\documentclass[reqno]{amsart}
\usepackage{graphicx}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 120, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/120\hfil Existence and uniqueness]
{Solutions of fourth-order partial differential equations in  a
noise removal model}

\author[Q. Liu, Z. Yao, Y. Ke \hfil EJDE-2007/120\hfilneg]
{Qiang Liu, Zhengan Yao, Yuanyuan Ke}  

\address{Qiang Liu \newline
Department of Mathematics, Jilin University, Changchun 130012,
China} 
\email{matliu@126.com}

\address{Zhengan Yao \newline
Department of Mathematics, Sun Yat-Sen University, Guangzhou 510275,
China} 
\email{mcsyao@mail.sysu.edu.cn}

\address{Yuanyuan Ke \newline
Department of Mathematics, Jilin University, Changchun 130012,
China} 
\email{keyy@jlu.edu.cn (corresponding author)}


\thanks{Submitted April 10, 2007. Published September 14, 2007.}
\thanks{Supported by grants NNSFC-10531040, NNSFC-10471156, NSFGD-4009793
 and \hfill\break\indent NSFGD-06300481}
\subjclass[2000]{35K65, 35M10} 
\keywords{Existence; uniqueness; fourth-order; noise removal}


\begin{abstract}
In this paper, we discuss the existence and uniqueness of weak
solutions for a fourth-order partial differential equation stemmed
from image processing for noise removal. We also present some
numerical tests for high order filters.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}


\section{Introduction}
We study the  fourth-order initial-boundary value problem
\begin{gather}
\frac{\partial u}{\partial t} + \frac{\partial ^2}{\partial x^2}
    \Phi'\big(\frac{\partial ^2 u}{\partial x^2}\big)= 0
    \quad(x,t) \in Q_T, \label{tv1}\\
u(0,t) = u(1,t) = u'(0,t) = u'(1,t) = 0 \quad t\in(0,T),\label{bd1}\\
u(x,0) = u_0(x)\quad x\in I \label{bd3},
\end{gather}
where $I = (0,1)$, $Q_T = I \times (0,T)$ and $\Phi: \mathbb{R} \to
\mathbb{R^+}$ is an $N$ function; i.e. $\Phi(\cdot)$ is even,
continuous, convex with $\Phi>0$  for $t>0$,
\begin{equation}\label{cd1}
\lim_{t\to 0}\frac{\Phi(t)}{t} \to 0 \quad\text{and}\quad \lim_{t\to
\pm\infty}\frac{\Phi(t)}{|\,t|} \to + \infty.
\end{equation}
Here we assume that $\Phi$ satisfies the $\Delta_2 $-condition:
\begin{equation}\label{cd2}
\Phi(2\xi) \le K\Phi(\xi),\quad |\xi| \ge R,
\end{equation}
where $K>2$ and $R$ are two positive constants.

In recent years, many nonlinear PDEs are proposed to deal with the
trade-off between noise removal and edge preservation. Among them,
the fourth-order parabolic PDEs have drawn great interest
\cite{AP,CMM,ke,tai,you,XZ,XZ2}. Since they seek to minimize a cost
functional which is an increasing function of the absolute value of
the Laplacian of the image intensity function, they could decrease
the staircasing property which may be undesirable under some
circumstances \cite{AP,DSTC}. In general, the forms of fourth-order
PDEs are analogous with the second order ones. For example, in
\cite{you}, You and Kaveh proposed equation
$$
u_t = -\Delta(g(\Delta u)\Delta u),
$$
where $g(s) = {1}/{(1+s^2)}$, which is analogous with the
Perona-Malik model \cite{pm}. In \cite{tai}, Lysaker \emph{et al}
used the equation
$$
u_t = -\Delta\big(\frac{\Delta u}{|\Delta u|}\big),
$$
which is similar to TV model \cite{ROF92}. In \cite{SD}, Didas used
the equation \eqref{tv1} with $\Phi(x) = 2\lambda
\left(\sqrt{\lambda^2 - x^2}-\lambda \right)$, where $\lambda > 0$
and it is the Charbonnier filter \cite{GA}.


Our model includes a class of more general equations \cite{GA,SD},
e.g. $\Phi(s) = \frac 1p |s|^p$, $p>1$. When $p=2$, a linear filter
could be obtained. While this filter has very strong isotropic
smoothing properties and does not preserve edges very well. One
should then decrease $p$ in order to preserve the edges as much as
possible, that is to say fast diffusion is desired. There are some
other functions which satisfy the conditions \eqref{cd1} and
\eqref{cd2}, for example:
$$
\Phi(s) = |s|\ln(1+|s|),
$$
and
$$
\Phi(s) = |s|L_k(|s|),
$$
where $L_i(s) = \ln(1+L_{i-1}(s))$ $(i= 1,2,\dots,k)$ and $L_0(s)=
\ln (1+|s|)$, see \cite{FM,WZ}.


Although the effectiveness of fourth order diffusion equations for
noise removal has been proposed in \cite{AP,CH,CMM,ke,tai,you}, very
little has been known about theoretical analysis. We refer to
\cite{wzyl}, Chapter 4 for a nonlinear equation with
double-degeneracy, \cite{travel} for traveling wave solutions in one
dimension, \cite{BG} for the existence and uniqueness of \eqref{tv1}
for $\Phi^\prime (s) = \arctan(s)$, \cite{ke} for the existence of a
fourth order PDE by variational methods and \cite{XZ} for a
generalized thin film equation.


It is worth while mentioning that the initial data is chosen by the
original image generally. We take the zero boundary value conditions
for convenience, which corresponds to padding the boundary of the
image with black.

The plan of the paper is the following. In Section 2, we state some
preliminaries and the main theorem. Section 3 is devoted to the
proofs of our main results and Section 4 deals with some numerical
experiments using finite difference methods by an explicit scheme.


\section{Preliminaries and  Main Result}
In the following sections we always assume $\Phi(\cdot)$ is a
function satisfied the condition \eqref{cd1} and \eqref{cd2}. Then
the N-function $\Psi(\cdot)$ which conjugates to $\Phi(\cdot)$ is
defined by
$$
\Psi(s) = \sup_{t \in \mathbb{R}}\{t\cdot s - \Phi(t)\}.
$$
We have the following Young's inequality,
$$
s\cdot t \le \Phi(s) + \Psi(t).
$$
For all $|s|>R$, we get (see \cite{Ad,WZ})
\begin{equation}\label{ine1}
\Phi(s) \le \Phi'(s) s \le (K-1)\Phi(s)
\end{equation}
and
\begin{equation}\label{ine2}
0 \le \Psi(\Phi'(s)) = \Phi'(s)  s - \Phi(s) \le (K-2)\Phi(s).
\end{equation}
For any $s,t \in \mathbb{R}$, we have (see \cite{Ad,WZ})
\begin{equation}\label{ine3}
\left(\Phi'(s) -\Phi'(t)\right)\cdot(s-t) \ge 0.
\end{equation}


\begin{lemma}[\cite{Ad,WZ}] \label{con0}
If $\Psi$ conjugates to $\Phi$, then there exist positive numbers
$p>1$, $R>0$, $R'>0$, $K_1>0$ and $K_2>0$ such that for all $s,t \in
\mathbb{R}$,
\begin{gather}
\Phi(s) \le K_1 |s|^p, \quad |s| \ge R, \label{ine4} \\
\Psi(t) \ge K_2 |t|^{p'}, \quad |t| \ge R', \quad p' = \frac
p{p-1}.\label{ine5}
\end{gather}
\end{lemma}



\begin{lemma}[\cite{AF,WZ}] \label{con}
Suppose $\{f_j\}\subset L^1(I;\mathbb{R})$ satisfies that
$$
\int_I \Phi(f_j)dx \le C,
$$
where $C$ is a positive constant. Then there exist a subsequence
$\{f_{m_j}\}\subset\{f_j\}$ and a function $f\in L^1(I;\mathbb{R})$
such that
$$
f_{m_j} \rightharpoonup f \quad\text{weakly in $L^1(I,\mathbb{R})$
as $j\to\infty$}
$$
with
$$
\int_I \Phi(f) dx \le \liminf_{j\to\infty} \int_I \Phi(f_{m_j}) dx
\le C.
$$
\end{lemma}

Now we define the weak solution of problem \eqref{tv1}--\eqref{bd3}.
\begin{definition} \label{def1} \rm
Let $T$ be a fixed positive constant. A function $u: Q_T \to
\mathbb{R}$ is called a weak solution of the problem
\eqref{tv1}--\eqref{bd3}, if the following conditions are fulfilled:
\begin{itemize}
\item[(1)] $u\in C([0,T];L^2(I)) \cap L^{\infty}(0,T; W_0^{2,1}(I))$ and
$\iint_{Q_T} \Phi\big(\frac{\partial^2 u}{\partial x^2}\big)\,
dx\,dt < +\infty$.
\item[(2)] For any $\varphi \in C_0^{\infty}(Q_T)$,
$$
\iint_{Q_T} \big\{-u\frac{\partial \varphi}{\partial t} +
\Phi'\big(\frac{\partial^2 u}{\partial x^2}\big)\frac{\partial^2
\varphi}{\partial x^2}\big\} dx\,dt = 0.
$$
\item[(3)]
$ u(x,0) = u_0(x)$ in $L^2(I)$.
\end{itemize}
\end{definition}


We state our main result as follows.
\begin{theorem} \label{theorem1}
Let $u_0 \in L^2(I)$ with $\int_{I} \Phi(\frac{\partial^2
u_0}{\partial x^2})\, dx \le C$ and compatibility conditions on
$\{0,1\}\times \{t=0\}$. Then problem \eqref{tv1}--\eqref{bd3}
admits one and only one weak solution.
\end{theorem}

\section{Proof of the Main Theorem}
We use the time discrete method to construct an approximate
solution. Divide the interval $(0,T)$ into $N$ equal segments and
denote $h=T/N$. Consider the problem:
\begin{gather}
\frac 1h\left( u_{k+1} - u_{k} \right) + \frac{d^2}{dx^2}
\Phi'\big(\frac{d^2 u_{k+1}}{dx^2}\big) = 0, \label{lisan}\\
u_{k+1}(0) = u_{k+1}(1) = u'_{k+1}(0) = u'_{k+1}(1) = 0,\label{lbd1}
\end{gather}
where $k = 0,1,\dots, N-1$, and $u_0$ is the initial data.

\begin{lemma}\label{lemlisan}
For $u_k \in L^2(I)$, the problem \eqref{lisan}-\eqref{lbd1} admits
one and only one weak solution $u_{k+1} \in W_0^{2,1}(I)$, such that
for any $\phi(x)\in C_0^{\infty}(I)$,
\begin{equation}
\frac 1h\int_0^1\left(u_{k+1}-u_k\right)\phi dx + \int_0^1
\Phi'\big(\frac{d ^2 u_{k+1}}{d x^2}\big) \frac{d ^2 \phi}{d x^2} dx
= 0, \label{lwas}
\end{equation}
and
$$
\int_0^1 \Phi\big(\frac{d^2 u_{k+1}}{dx^2}\big) dx \le C,
$$
where $C$ is a constant depended only on $\|u_k\|_{L^2(I)}$ and $h$.
\end{lemma}

\begin{proof} We investigate the functional defined on
$W_0^{2,1}(I)$ by
\begin{equation*}
E (v) = \frac 1{2h}\int_0^1 (v - u_k)^2 dx + \int_0^1 \Phi\big(\frac
{d^2v}{dx^2}\big) dx.
\end{equation*}
We choose $v = 0$, then
$$
0\le \inf_{v\in W_0^{2,1}(I)} E(v) \le E(0) = \frac 1{2h}\int_0^1
u_k ^2 dx.
$$
By lemma \ref{con}, we can extract a minimizing sequence
$\{v_n\}_{n=1}^\infty \subset W_0^{2,1}(I)$ such that
$$
E(v_n) \to \inf_{v\in W_0^{2,1}(I)} E(v),\quad \text{as
$n\to\infty$},
$$
and
$$
\int_0^1 |v_n|^2 dx + \int_0^1 \Phi \big(\frac{d^2 v_n}{dx^2}\big)
dx \le C.
$$
By \eqref{cd1} and Lemma \ref{con}, we may find a subsequence
$\{v_{n_j}\}_{j=1}^\infty\subset\{v_n\}_{n=1}^{\infty}$ and a
function $u_{k+1}$, such that $v_{n_j} \rightharpoonup u_{k+1}$
weakly in $W_0^{2,1}(I)$ and
$$
\int_0^1 \Phi\big(\frac{d^2u_{k+1}}{dx^2}\big) \le C.
$$
Since $\Phi(s)$ is convex and by relaxation, we have that $u_{k+1}$
is a weak solution of the problem \eqref{lisan}--\eqref{lbd1}.

Assume $u_{k+1}$ and $v_{k+1}$ are both solutions of the problem
\eqref{lisan}--\eqref{lbd1}. Then for every $\phi(x)\in
C_0^{\infty}(I)$, we have
\begin{equation*}
\frac 1h\int_0^1(u_{k+1}-v_{k+1})\phi dx + \int_0^1\Big(
\Phi'\big(\frac{d ^2 u_{k+1}}{d x^2}\big) -\Phi'\big(\frac{d ^2
v_{k+1}}{d x^2}\big)\Big)\frac{d ^2 \phi}{d x^2} dx = 0.
\end{equation*}
By \eqref{ine2} and the approximation argument, we could take
$\phi(x) = u_{k+1} - v_{k+1}$ as the test function. We get
\begin{align*}
&\frac 1h\int_0^1\left(u_{k+1}-v_{k+1}\right)^2 dx \\
&+ \int_0^1\Big( \Phi'\big(\frac{d ^2 u_{k+1}}{d x^2}\big)
-\Phi'\big(\frac{d ^2 v_{k+1}}{d x^2}\big)\Big) \Big( \frac{d ^2
u_{k+1}}{d x^2}- \frac{d ^2 v_{k+1}}{d x^2}\Big)dx = 0.
\end{align*}

By \eqref{ine3}, the two terms on the left hand side are both
nonnegative. We get $u_{k+1} = v_{k+1}$ a.e. in $I$. Then the proof
is complete.
\end{proof}


Let $\chi^{h,j}(t)$ be the indicator function of $[h(j-1),hj)$. We
construct an approximate function by
\[
u^h(x,t) = \sum_{j=1}^{N}\chi^{h,j}(t)u_{j-1}(x) \quad \text{with}\quad
u^h(x,0) = u_{0}(x).
\]

\begin{lemma}\label{lmgu1}
For the weak solution $u_{k+1}$ of the problem
\eqref{lisan}--\eqref{lbd1}, the following estimates hold
\begin{gather}\label{es1}
h\sum_{k=0}^{N-1} \int_0^1
    \Phi'\left(\frac{d^2u_{k+1}}{dx^2}\right)\frac{d^2u_{k+1}}{dx^2} dx
    \le C, \\
\label{es2} \sup_{0<t<T}\int_0^1 \Phi\left(\frac{\partial
    ^2u^h}{\partial x^2}\right) dx \le C,
\end{gather}
where $C$ is a constant independent of $h$.
\end{lemma}

\begin{proof}
Noticing that $C_0^{\infty}(I)$ is dense in $W^{2,1}_0(I)$, we may
choose $\phi(x)\in W^{2,1}_0(I)$ as the test function in
\eqref{lwas}. Let $\phi(x) =u_{k+1}$ in \eqref{lwas}. Then
$$
\frac 1h\int_0^1 (u_{k+1}-u_{k})u_{k+1} dx + \int_0^1
\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2} dx =
0.
$$
So we have
\begin{align*}
\frac 1h\int_0^1 u_{k+1}^2 dx + \int_0^1
\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2} dx \le
\frac {1}{2h}\int_0^1 (u_{k+1}^2 + u_{k}^2) dx ;
\end{align*}
i.e.,
\begin{align}\label{tmp1}
\frac 12 \int_0^1 u_{k+1}^2 dx + h\int_0^1
\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2} dx \le
\frac {1}{2}\int_0^1 u_{k}^2 dx.
\end{align}
Summing up \eqref{tmp1} for $k$ from $0$ to $N-1$, we have
$$
\frac 12 \int_0^1 u_N^2 dx + h\sum_{k=0}^{N-1}\int_0^1
\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2} dx \le
\frac {1}{2}\int_0^1 u_{0}^2 dx.
$$
Then \eqref{es1} is obtained.

Letting $\phi(x) = u_{k+1}-u_{k}$ in \eqref{lwas}, we obtain
$$
\frac 1h \int_0^1 (u_{k+1} - u_{k})^2 dx + \int_0^1
\Phi'\big(\frac{d^2 u_{k+1}}{dx^2}\big)\big(\frac{d^2
u_{k+1}}{dx^2}-\frac{d^2u_{k}}{dx^2}\big) dx = 0.
$$
Since the first term of above equality is nonnegative, by Young's
inequality and \eqref{ine2}, we have
\begin{align*}
&\quad\int_0^1
    \Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2}
    dx \\
&\le \int_0^1 \Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)
    \frac{d^2u_{k}}{dx^2}dx \\
&\le \int \Psi\Big(\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\Big) dx +
    \int_0^1 \Phi\big(\frac{d^2u_{k}}{dx^2}\big) dx \\
& = \int_0^1
    \Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_{k+1}}{dx^2}
    dx - \int_0^1\Phi\big(\frac{d^2u_{k+1}}{dx^2}\big) dx +
    \int_0^1 \Phi\big(\frac{d^2u_{k}}{dx^2}\big) dx.
\end{align*}
Thus
$$
\int_0^1 \Phi\big(\frac{d^2u_{k+1}}{dx^2}\big) dx \le \int_0^1
\Phi\big(\frac{d^2u_{k}}{dx^2}\big) dx.
$$

For any $m$ with $1\le m \le N-1$, summing up the above inequality
for $k$ from $0$ to $m-1$, we have
$$
\int_0^1 \Phi\big(\frac{d^2u_{m}}{dx^2}\big) dx \le \int_0^1
\Phi\big(\frac{d^2u_{0}}{dx^2}\big) dx.
$$
So we get \eqref{es2} and the proof is complete.
\end{proof}


\begin{lemma}\label{lem55}
For the weak solution $u_{k+1}$ of \eqref{lisan}--\eqref{lbd1}, we
have
\begin{equation}\label{lem5}
-Ch \le \int_0^1 |u_{k+1}|^2 - |u_k|^2 dx \le 0,
\end{equation}
where $C$ is a positive constant independent of $h$.
\end{lemma}

\begin{proof}
The second inequality of \eqref{lem5} is obvious by \eqref{tmp1}.
Choosing $\phi(x) = u_k$ in \eqref{lwas}, we have
$$
\frac 1h \int_0^1(u_{k+1}-u_k)u_k dx + \int_0^1
\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_k}{dx^2} dx = 0.
$$
So by Young' inequality and inequality \eqref{ine2}, we have
\begin{align*}
\frac 1h \int_0^1(u_{k}-u_{k+1})u_k dx \le& \int_0^1
    \Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\frac{d^2u_k}{dx^2}
    dx\\
\le& \int_0^1 \Psi\Big(\Phi'\big(\frac{d^2u_{k+1}}{dx^2}\big)\Big)dx
    +\int_0^1 \Phi\big(\frac{d^2u_{k}}{dx^2}\big)dx\\
    \le& (K-2)\int_0^1 \Phi\big(\frac{d^2u_{k+1}}{dx^2}\big)dx +
    \int_0^1 \Phi\big(\frac{d^2u_{k}}{dx^2}\big)dx.
\end{align*}
By  \eqref{es2} of Lemma \ref{lmgu1}, we have
$$
\int_0^1 u_k^2 dx - \int_0^1 u_{k+1}u_k dx \le Ch.
$$
Therefore,
$$
\int_0^1 u_k^2 dx  \le Ch + \int_0^1 u_{k+1}u_k dx \le Ch + \frac 12
\int_0^1 u_k^2 dx + \frac 12 \int_0^1 u_{k+1}^2 dx.
$$
Thus, we obtain that
$$
\frac 12\int_0^1 u_k^2 dx  \le Ch + \frac 12 \int_0^1 u_{k+1}^2 dx.
$$
So the proof of this lemma is complete.
\end{proof}


\begin{corollary}\label{coro1}
\begin{equation*}
\int_0^1 |u^h|^2 dx \le \int_0^1 |u_0|^2 dx.
\end{equation*}
\end{corollary}

\begin{proof}[Proof of Theorem \ref{theorem1}]
Let
$$
\xi_h = \Phi'\big(\frac{\partial^2 u^h}{\partial x^2}\big)
\quad\text{and}\quad \Delta^h u^h = u_{k+1}-u_{k}.
$$
By \eqref{lwas} we see that
\begin{equation}\label{liss}
\iint_{Q_T}\left(\frac 1h \Delta^h u^h \varphi + \xi_h
\frac{\partial ^ 2 \varphi}{\partial x^2}\right) dx\,dt= 0,
\end{equation}
for any $\varphi \in C_0^\infty(Q_T)$.

By Lemma \ref{con}, Lemma \ref{lmgu1}  and Corollary \ref{coro1}, we
can draw a subsequence $\{u^h\}$, denoted still by $\{u^h\}$, such
that
\begin{gather*}
u^h \rightharpoonup u \quad\text{weakly * in
    $L^\infty(0,T,W_0^{2,1}(I))$},\\
\iint_{Q_T}\Phi\big(\frac{\partial^2 u}{\partial x^2}\big)dx\,dt \le
    C,\\
u^h \rightharpoonup u \quad\text{weakly * in
$L^\infty(0,T,L^{2}(I))$}.
\end{gather*}
By  \eqref{ine2},
$$
\iint_{Q_T} \Psi\left(\xi_h\right) dx\,dt \le \iint_{Q_T}
(K-2)\Phi\big(\frac{\partial u^h}{\partial x^2}\big) dx\,dt \le C.
$$
And by lemma \ref{con0},
$$
\iint_{Q_T} |\xi_h|^{p'} dx\,dt \le C,
$$
for some $p'>1$. Thus, we may extract a subsequence from $\xi_h$,
denoted still by $\xi_h$, such that
$$
\xi_h \rightharpoonup \xi \quad \text{weakly in $L^{p'}(\Omega)$}.
$$
Since $\Psi(s)$ is a convex function, we obtain
$$
\iint_{Q_T} \Psi(\xi) dx\,dt \le \liminf_{h\to 0}
\iint_{Q_T}\Psi(\xi_h) dx\,dt \le C.
$$
Using Young's inequality again, we have
$$
\iint_{Q_T}\big|\xi\cdot \frac{\partial ^2 u}{\partial x^2}\big|
dx\,dt \le \iint_{Q_T} \Psi(\xi) + \Phi\big(\frac{\partial ^2
u}{\partial x^2}\big)dx\,dt \le C.
$$
By the discrete equation \eqref{liss}, we see that
$$
\frac 1h \Delta^h u^h \quad\text{is bounded in
$L^\infty(0,T;W^{-2,1}(I))$}
$$
and
$$
\frac 1h \Delta^h u^h \rightharpoonup \frac{\partial u}{\partial t}
\quad\text{weakly * in $L^\infty(0,T;W^{-2,1}(I))$}.
$$
Letting $h\to 0$ in \eqref{liss}, we have in the sense of
distributions
\begin{equation}\label{dis}
\frac{\partial u}{\partial t} + \frac{\partial^2 \xi}{\partial x^2}
= 0.
\end{equation}

Now we will prove $\xi = \Phi'\big(\frac{\partial^2 u}{\partial
x^2}\big)$. Denote
$$
f_h(t) = \frac{t-kh}{2h}\Big(\int_0^1 |u_{k+1}|^2dx - \int_0^1
|u_k|^2dx\Big) + \frac 12 \int_0^1 u_k^2dx,
$$
where $kh < t \le (k+1)h$, $k=0,1,2,\dots, N-1$. By \eqref{lem5}, we
have
\begin{gather*}
\frac 12 \int_0^1 |u_k|^2 dx - Ch \le f_h(t) \le \frac 12 \int_0^1 |u_k|^2dx,\\
-C \le f'_h(t) \le 0.
\end{gather*}
According to the Ascoli-Arzela theorem, there exists a function
$f(t) \in C([0,T])$, such that
$$
\lim_{h\to 0}f_h(t) = \frac 12\lim_{h\to 0}\int_0^1 |u^h|^2 dx =
f(t) \quad\text{uniformly for $t\in[0,T]$}.
$$
It follows from \eqref{tmp1} that
$$
\frac 12\int_0^1 |u^h|^2 dx + \iint_{Q_T} \Phi'\big(\frac{\partial
^2 u^h}{\partial x^2}\big)\frac{\partial ^2 u^h}{\partial x^2}
dx\,dt \le \frac 12 \int_0^1 |u_0|^2 dx.
$$
Letting $h\to 0 $ in the above inequality we have
\begin{align*}
&\liminf_{h\to 0} \iint_{Q_T} \Phi'\left(\frac{\partial ^2
    u^h}{\partial x^2}\right)
    \frac{\partial ^2 u^h}{\partial x^2} dx\,dt \\
&\le  f(0)- f(T) \\
&= \lim_{\varepsilon\to 0^+} \frac 1\varepsilon
    \int_0^{T-\varepsilon} (f(t)-f(t+\varepsilon)) dt \\
&= \lim_{\varepsilon\to 0^+} \lim_{h \to 0} \frac 1{2\varepsilon}
    \int_0^{T-\varepsilon} \int_0^1 (|u^h(x,t)|^2 - |u^h(x,t+\varepsilon)|^2) dx\,dt \\
&\le \lim_{\varepsilon\to 0^+} \frac 1\varepsilon
    \int_0^{T-\varepsilon}\int_0^1 (u(x,t)-u(x,t+\varepsilon))\cdot u\,dx\,dt\\
&\le  -\int_0^T \langle \frac{\partial u}{\partial t},u \rangle dt,
\end{align*}
where $\langle \cdot\rangle $ denotes the dual product of the
function in $W^{-2,1}(I)$ and $W_0^{2,1}(I)$. So we have

\begin{equation}\label{lsc}
\liminf_{h\to 0} \iint_{Q_T} \Phi'\big(\frac{\partial ^2
u^h}{\partial x^2}\big)\frac{\partial ^2 u^h}{\partial x^2} dx\,dt
\le \iint_{Q_T} \xi \frac{\partial^2 u}{\partial x^2} dx\,dt.
\end{equation}

Define $F[u] = \int_0^1 \Phi\big(\frac{\partial^2 u}{\partial
x^2}\big) dx$ and choose a function $g\in L^\infty(0,T;
W_0^{2,1}(I))$ with $\iint_{Q_T} \Phi\big(\frac{\partial^2
g}{\partial x^2}\big)\, dx\,dt < +\infty$. Because $\Phi(s)$ is
convex, we have
$$
\iint_{Q_T} \Phi\big(\frac{\partial^2 g}{\partial x^2}\big) dx\,dt -
\iint_{Q_T} \Phi\big(\frac{\partial^2 u^h}{\partial x^2}\big)dx\,dt
\ge \iint_{Q_T}\Phi'\big(\frac{\partial^2 u^h}{\partial x^2}\big)
\frac{\partial^2 (g -u^h)}{\partial x^2} dx\,dt.
$$
Letting $h\to 0$ and by \eqref{lsc}, we get
$$
\iint_{Q_T} \Phi\big(\frac{\partial^2 g}{\partial x^2}\big) dx\,dt -
\iint_{Q_T} \Phi\big(\frac{\partial^2 u}{\partial x^2}\big) dx\,dt
\ge \iint_{Q_T} \xi\cdot\frac{\partial^2 (g -u)}{\partial x^2}
dx\,dt.
$$
Replacing $g$ by $\varepsilon g + u$, we see that
$$
\frac 1\varepsilon(F[u+\varepsilon g]-F[u]) \ge \iint_{Q_T} \xi\cdot
\frac{\partial^2 g}{\partial x^2}dx\,dt
$$
and
$$
\iint_{Q_T} \frac{\delta F[u]}{\delta u} g dx\,dt = \iint_{Q_T}
\Phi'\big(\frac{\partial^2 u}{\partial x^2}\big) \frac{\partial ^2
g}{\partial x^2} dx\,dt \ge \iint_{Q_T} \xi \cdot \frac{\partial ^2
g}{\partial x^2} dx\,dt.
$$
Due to the arbitrariness of $g$, we get that $\xi =
\Phi'\big(\frac{\partial^2 u}{\partial x^2}\big)$. By \eqref{dis},
$u$ is the weak solution of the problem \eqref{tv1}--\eqref{bd3}.


Next, we  prove the uniqueness of the weak solution of the problem
\eqref{tv1}--\eqref{bd3}. Suppose there exist two weak solutions $u$
and $v$. Using an approximation technique (see \cite{wzyl,XZ}), for
any test function $\varphi(x,t)\in C^{\infty}(\bar Q_T)$, we have
$$
\iint_{Q_T} -(u-v)\frac{\partial \varphi}{\partial t} dx\,dt +
\iint_{Q_T} \Big(\Phi'\big(\frac{\partial^2 u}{\partial x^2}\big) -
\Phi'\big(\frac{\partial^2 v}{\partial
x^2}\big)\Big)\frac{\partial^2 \varphi}{\partial x^2} dx\,dt = 0.
$$
Furthermore, we may take $u-v$ as a test function and then get
$$
\frac 12\int_0^1 |u-v|^2(t) dx\,dt + \iint_{Q_t}
\Big(\Phi'\big(\frac{\partial^2 u}{\partial x^2}\big) -
\Phi'\big(\frac{\partial^2 v}{\partial
x^2}\big)\Big)\Big(\frac{\partial^2 u}{\partial x^2} -
\frac{\partial^2 v}{\partial x^2} \Big) dx\,dt = 0,
$$
where $Q_t = (0,t)\times I$. Since the two terms on the left hand
side are nonnegative by inequality \eqref{ine4}, we have $u=v$ a.e.
in $Q_T$. Thus the proof is complete.
\end{proof}


\section{Numerical experiments}
After the theoretical analysis, we shall do some numerical tests of
higher order filters in practice to compare our model with the other
well-known models of \cite{pm,you}. In our model, we take $ \Phi(s)
= |s|\ln(1+|s|)$. For convenience, we are in favor of implementation
of an explicit Euler method, i.e.
$$
\frac{u^{k+1} - u^k}{\Delta t} + \frac{\partial ^2}{\partial x^2}
\Phi'\big(\frac{\partial ^2 u^k}{\partial x^2}\big)= 0, \quad(x,t)
\in Q_T.
$$
For each figure, we use $1$ for space steps, $0.2$ for time steps of
figure (c) and  $0.001$ for time steps of figures (d) and (e).
Steady state was achieved for figure (d) and figure (e) in less than
$20000$ iterations. In figure (c), we fixed the number of iterations
to $1500$.

Fig.1 (a) shows the initial signal and (b) the noisy signal. By the
figures from (c) to (e), we could conclude that the second order
filtering yields enhancement of edges and staircase-like structures,
the fourth order filtering results tend to be piecewise linear with
enhanced curvature. At the same time we could also see that the
fourth order filtering is further affirmed by the almost piecewise
constant derivative which is also shown in Figure 1.



\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.45\textwidth]{fig1a}
\includegraphics[width=0.45\textwidth]{fig1b}  \\
\includegraphics[width=0.45\textwidth]{fig1c}
\includegraphics[width=0.45\textwidth]{fig1d} \\
\includegraphics[width=0.45\textwidth]{fig1e}
\end{center}
\caption{One-dimensional signal evaluation: original signal,  noisy
signal, and restored by second order Perona-Malik model,
fourth-order Perona-Malik model and our model.}
\end{figure}

\subsection*{Acknowledgements}
The authors would like to express their sincerely thanks to Prof.
J.X. Yin for the advised discussing; also to Dr. M. Xu for providing
important references for this paper. The authors would  like to
thank the anonymous referees for their valuable suggestions for the
revision of the manuscript.


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\end{document}
